Diffeomorphism from $\mathbb{R}/2\pi\mathbb{Z}$ to $S^1$ I'm struggling to prove that the map $F: \mathbb{R}/2\pi\mathbb{Z} \to S^1, \; F([t]) = e^{it}$ is a diffeomorphism. I think what I'm missing is the right choice of atlases for the two manifolds.
This is my attempt: let
\begin{cases}
U_1 = \{ e^{it} \mid t \in (-\pi, \pi) \}, \; \phi_1(e^{it}) = t\\
U_2 = \{ e^{it} \mid t \in (0, 2\pi) \}, \; \phi_2(e^{it}) = t
\end{cases}
Then, the collection $\{(U_i, \phi_i)\}$ is an atlas for $S^1$.
We have that for every $x, y \in \mathbb{R}$, $[x] = [y] \iff x-y $ is an integer multiple of $2\pi$. Also, for every $x \in \mathbb{R} \; \exists! \, y_x \in [0, 2\pi) \, | \, x \sim y_x $.
Let $\psi: \mathbb{R}/2\pi\mathbb{Z} \to [0, 2\pi), \, \psi([x]) = y_x$. Then, ${(\mathbb{R}/2\pi\mathbb{Z}, \psi)}$ is an atlas for $\mathbb{R}/2\pi\mathbb{Z}$.
These atlases however do not seem to work because when I compose $F^{-1}$ with the charts I do not get continuous functions. Can anyone help me?
 A: Let $\tilde{F}:\mathbb{R}\rightarrow S^{1}$ be given by $\tilde{F}(t) = e^{it}$. Here we identify $S^{1}$ with the set of points in $\mathbb{C}$ given by $||z||=1$. We make no use of the complex structure other that in defining the map and use the fact that $\tilde{F}(t)=(\cos(t),\sin(t))$, when thought of as a map to $\mathbb{R}^{2}$.
First, $\tilde{F}:\mathbb{R}\rightarrow S^{1}$ is a local diffeomorphism. To see this, we check the rank of $dF$ is never zero:
$$\begin{align}
d\tilde{F}(\partial_{t}) 
 &= \frac{\partial \tilde{F}_{1}}{\partial t}\partial x+\frac{\partial \tilde{F}_{2}}{\partial t}\partial y\\
&=-\sin(t)\partial x +\cos(t)\partial y
\end{align}$$
Since $\cos(t)$ and $\sin(t)$ never vanish together, the rank of $dF$ is alway 1. By the inverse function theorem, $\tilde{F}$ is a local diffeomorphism.
Next, we consider the quotient. In general, a quotient space obtained from a manifold is generally not a manifold. However, if we consider any ball of radius $r<\pi$, the quotient map $q:\mathbb{R}\rightarrow \mathbb{R}/2\pi\mathbb{Z}$ restricts to a homeomorphism: for $x,y\in B_{r}(z)$, $|x-y|< 2\pi$. Thus, $q\big|_{B_{r}(z)}:B_{r}(z)\rightarrow [B_{r}(z)]$ bijectively. Further, $ q^{-1}([B_{r}(z)]) = \bigcup (k+B_{r}(z))$,  for $k\in\mathbb{Z}$ and hence [B_{r}(z)] is open in $\mathbb{R}/2\pi\mathbb{Z}$. This establishes that $q\big|_{B_{r}(z)}:B_{r}(z)\rightarrow [B_{r}(z)]$ is a continuous bijective open map, and hence a homeomorphism. Open sets of these type cover $\mathbb{R}/2\pi\mathbb{Z}$, So they can be used to give a smooth structure on $\mathbb{R}/2\pi\mathbb{Z}$ (The transition functions are translation by some integer which is smooth). By construction these local homeomorphisms are local diffeomorphism (we are using the smooth structure on $\mathbb{R}/2\pi\mathbb{Z}$ obtained from these functions making them smooth maps).
Finally, if we look at $\tilde{F}$ and the induced map $F:\mathbb{R}/2\pi\mathbb{Z}\rightarrow S^{1}$. We need to see this is well defined: $\tilde{F}^{-1}(p) =\{t + 2\pi k| k\in \mathbb{Z}\}$ for $p=\tilde{F}(t)$. Since all these points are in the same equivalence class by definition, $\tilde{F}$ is constant on $q^{-1}([t])$. So we have a well defined continuous map $F$. Now this map is bijective since $q^{-1}([t])=\tilde{F}^{-1}(p)$ for any $t\in \mathbb{R}$ such that $\tilde{F}(t)=p$. Further, by restricting to a chart constructed above, we have that $F\circ q^{-1}\big|_{[b_{r}(z)]}= \tilde{F}\big|_{b_{r}(z)}$ which is a local diffeomorphism. Hence $F$ is a bijective local diffeomorphism, hence a global diffeomorphism.
